Understanding Elementary Shapes - Class 6 Maths - Extra Questions

(i) Area of $$\Delta ABC = {1 \over 2}BC \times AC$$

$$= {1 \over 2} \times a \times b = {{ab} \over 2}$$ 3

$$area = {1 \over 2} \times CD \times AB$$

$${1 \over 2} \times p \times c = {{pc} \over 2}$$

$${{ab} \over 2} = {{pc} \over 2}$$

$$ab = pc......(1)$$

(ii) In $$\Delta ABC$$

$$A{B^2} = A{C^2} + B{C^2}$$

$${\left( {{{ab} \over p}} \right)^2} = {b^2} + {a^2}$$

$${1 \over {{p^2}}} = {{{b^2} + {a^2}} \over {{a^2} + {b^2}}}$$

$${1 \over {{p^2}}} = {1 \over {{a^2}}} + {1 \over {{b^2}}}$$

The sign $$\bot$$ in $$AE \bot DC$$ shows that $$AE$$ is perpendicular to $$DC$$ .

It is given that in $$\Delta DEF,$$   $$\angle D = 90^\circ$$

Since, one of the angles of the given triangle is a right angle, therefore, the given triangle is a right angled triangle.

It is given that in $$\Delta LMN$$ ,

$$\angle L = 30^\circ$$ , $$\angle M = 70^\circ$$ and $$\angle N = 80^\circ$$

Since, all the angles of the given triangle measures less than $$90^\circ$$ , therefore, the given triangle is an acute triangle.

We have,

$$ADE$$ is a right angled triangle.

Therefore ,By Pythagoras theorem,

$$A{{E}^{2}}+E{{D}^{2}}=A{{D}^{2}}$$

$$\Rightarrow A{{E}^{2}}=A{{D}^{2}}-E{{D}^{2}}\,\,......\,\left( 1 \right)$$

Again, $$In\,\Delta AEB$$

$$A{{B}^{2}}=A{{E}^{2}}+B{{E}^{2}}$$ ……. (2)

And now,

$$BE=BD-ED$$  …… (3)

Substituting equation (2) and (3) in (1), we get,

$$A{{B}^{2}}=A{{D}^{2}}-E{{D}^{2}}+{{\left( BD-ED \right)}^{2}}$$

$$=A{{D}^{2}}-E{{D}^{2}}+B{{D}^{2}}+E{{D}^{2}}-2BD\times ED$$

Now, $$BD=\dfrac{BC}{2}\,\,\,\,\,\,\,\,\,\,\,\left( \therefore \,D\,is\,\text{midpoint} \right)$$

Therefore,

$$A{{B}^{2}}=A{{D}^{2}}+B{{D}^{2}}-2BD\times ED$$

$$=A{{D}^{2}}+{{\left( \dfrac{BC}{2} \right)}^{2}}-2\left( \dfrac{BC}{2} \right)\times ED$$

$$=A{{D}^{2}}-BC\times ED+\dfrac{B{{C}^{2}}}{4}$$

Given that,

$$\angle ABC={{90}^{0}}\,\And \,\,BD\bot AC$$

and

  $$AB=5.7\,cm$$

  $$BD=3.8cm$$

  $$CD=4.4cm$$

$$BC=?$$

Let $$AD=x\,cm$$   and  $$BC=y\,cm$$

Then, $$In\,\Delta ABD$$

By pythagoras theorem,

  $$A{{B}^{2}}=A{{D}^{2}}+D{{B}^{2}}$$

  $${{\left( 5.7 \right)}^{2}}={{y}^{2}}+{{\left( 3.8 \right)}^{2}}$$

  $$32.49={{y}^{2}}+13.3$$

  $${{y}^{2}}=32.49-13.3$$

  $${{y}^{2}}=19.19$$

  $$y=4.38\,cm$$

Now, $$In\,\Delta ABC$$

By Pythagoras theorem,

  $$A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}$$

  $${{\left( y+4.4 \right)}^{2}}={{\left( 5.7 \right)}^{2}}+{{x}^{2}}$$

  $${{\left( 4.38+4.4 \right)}^{2}}=32.49+{{x}^{2}}$$

  $${{\left( 8.78 \right)}^{2}}-32.49={{x}^{2}}$$

  $${{x}^{2}}=77.87-32.49$$

  $${{x}^{2}}=45.38$$

  $$x=6.73\,cm$$

$$\\From \, the \, above \, figure$$

$$PR \cos60^0=6$$

$$\therefore\>PR=(\dfrac{6}{cos60^0})=12$$

$$QR=PR \sin60^0$$

$$=12\times(\dfrac{\sqrt{3}}{2})=6\sqrt{3}$$

We have,

$$X$$ divides $$PQ$$ in ratio $$1:2$$ .

$$\Rightarrow XQ=2PX$$

$$\Rightarrow XQ=\dfrac{2}{3}PQ\,\,......\,\,\left( 1 \right)$$

Also, $$Y$$ divides QR in ratio $$2:1$$

$$QY=2YR$$

$$\Rightarrow QY=\dfrac{2}{3}QR\,\,.......\,\,\left( 2 \right)$$

In $$\Delta \,PQY,$$

$$P{{Y}^{2}}=P{{Q}^{2}}+Q{{Y}^{2}}$$

$$\Rightarrow P{{Y}^{2}}=P{{Q}^{2}}+{{\left( \dfrac{2}{3}QR \right)}^{2}}$$

$$\Rightarrow P{{Y}^{2}}=P{{Q}^{2}}+\dfrac{4}{9}Q{{R}^{2}}$$

$$\Rightarrow 9P{{Y}^{2}}=P{{Q}^{2}}+4Q{{R}^{2}}\,\,.......\,\,\left( 3 \right)$$

In $$\Delta XQR,$$

So,

$$X{{R}^{2}}=X{{Q}^{2}}+Q{{R}^{2}}$$

$$\Rightarrow X{{R}^{2}}={{\left( \dfrac{2}{3}PQ \right)}^{2}}+Q{{R}^{2}}$$

$$\Rightarrow X{{R}^{2}}=\dfrac{4P{{Q}^{2}}}{9}+Q{{R}^{2}}$$

$$\Rightarrow X{{R}^{2}}=4P{{Q}^{2}}+9Q{{R}^{2}}\,\,......\,\,\left( 4 \right)$$

On adding ( 3) and ( 4) to

$$9P{{Y}^{2}}+9X{{R}^{2}}=9P{{Q}^{2}}+4Q{{R}^{2}}+4P{{Q}^{2}}+9Q{{R}^{2}}$$

$$\Rightarrow 9\left( P{{Y}^{2}}+X{{R}^{2}} \right)=13\left( P{{Q}^{2}}+Q{{R}^{2}} \right)$$

$$\Rightarrow 9\left( P{{Y}^{2}}+X{{R}^{2}} \right)=13P{{R}^{2}}$$

Hence, proved.

Length of hypotensue $$=3\sqrt{10}$$

Let the length of shorter side = $$x$$

and length of longer side = $$y$$

According to pythagoras theorem,

$${{x}^{2}}+{{y}^{2}}={{\left( 3\sqrt{10} \right)}^{2}}=90$$

$${{x}^{2}}+{{y}^{2}}=90\,\,......\,\,\left( 1 \right)$$

The smaller side is tripled $$=3x$$

The larger side is doubled $$=2y$$

New hypotenuse $$=9\sqrt{5}$$

Now,

$${{\left( 3x \right)}^{2}}+{{\left( 2y \right)}^{2}}={{\left( 9\sqrt{5} \right)}^{2}}$$

$$9{{x}^{2}}+4{{y}^{2}}=405$$

$$5{{x}^{2}}+4{{x}^{2}}+4{{y}^{2}}=405$$

$$5{{x}^{2}}+4\left( {{x}^{2}}+{{y}^{2}} \right)=405$$

$$5{{x}^{2}}+4\times 90=405$$

$$5{{x}^{2}}=405-360$$

$${{x}^{2}}=\dfrac{45}{5}$$

$${{x}^{2}}=9$$

$$x=3$$

Put $$x=3$$ in (1) and we get,

$$y=9$$

$$\textbf{Step -1: Checking dimensions of a solid.}$$

                  $$\text{A solid has length, breadth and height as dimensions.}$$

                  $$\therefore \text{ A solid has 3 dimensions.}$$

$$\textbf{Final Answer: A solid has 3 dimensions.}$$

Since $$\angle {MNL} = 90^0$$  and

Sum of two acute angle s,

$$\angle M + \angle L = 30^0 + 60^0$$

$$= 90^0$$

Therefore, it is a right-angled triangle

The least number of planes that can enclose a solid is 4.

The name of the solid is Tetrahedron.

Number of faces $$=8$$

Number of vertices $$=16$$

Number of edges $$=26$$

Using Euler’s formula

$$F+V-E=2$$

Checking LHS, 

$$8+16-26$$

$$8+16-26$$

$$-2\neq2$$

No, a polyhedron cannot have $$8$$ faces, $$26$$ edges and $$16$$ vertices.

It is given that in $$\Delta ABC$$ ,

$$AB = 8.7\;{\rm{cm}}$$ , $$AC = 7\;{\rm{cm}}$$ and $$BC = 6\;{\rm{cm}}$$ .

Since all the sides of the given triangle are different, therefore, the given triangle is scalene triangle.